Oblicz a)  [latex]log _{2} 48-log _{2}3= [/latex] b) [latex]log _{3} frac{1}{6} + log _{3} frac{2}{3} =[/latex] c)[latex]2log _{ frac{1}{4} 8-3log _{ sqrt{3}}9[/latex]   Blagam o szybka odpowiedz na egzamin :(

 [latex]a) \ log _{2} 48-log _{2}3= log_{2}frac{48}{3} = log_{2}16 =4 \ \2^x=16\ \2^x =2^4 \ \x=4[/latex] [latex]b) \ log _{3} frac{1}{6} + log _{3} frac{2}{3} = log_{3} (frac{1}{6}*frac{2}{3}) =log_{3} frac{1}{9} = -2 \ \ \3^x=frac{1}{9}\ \3^x=9 ^{-1} \ \3^x = (3^2)^{-1} \ \3^x = 3)^{-2} \ \x= -2[/latex] [latex]c)\ \ 2log _{ frac{1}{4}} 8-3log _{ sqrt{3}}9 =2*(- frac{3}{2}) - 3 *4 = -3-12 = - 15 \ \ log _{ frac{1}{4}} 8 = 1 \ \(frac{1}{4})^x =8 \ \(2 ^{-2})^x =2^3\ \2^{-2x} = 2^3 \ \-2x=3 / :2 \ \ x= -frac{3}{2}[/latex] [latex] log _{ sqrt{3}}9 = log _{ sqrt{3}}9 = 4 \ \ (sqrt{3})^x =9 \ \( 3^{frac{1}{2}})^x =9 \ \ 3^{frac{1}{2}x} =3^2\ \frac{1}{2}x = 2 /*2\ \x=4[/latex]